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Let $A,B \subseteq \mathbb{R}^d$ be non-empty sets. Define their distance to be

$$ d(A,B) = \inf \{ ||x-y|| : x \in A, \; \; y \in B \} $$

For any $A,B$, do we have that $d(A,B) = d( \overline{A}, \overline{B} ) $.

Is the following proof correct?

Proof

Note, you always have $d(A,B)\geq d(\bar{A},\bar{B})$ since the supremum is taken over a bigger set on the right hand side.

If $d(\bar{A},\bar{B}) = d$ then $\exists x_1,x_2,...\in \bar{A}$ and $y_1,y_2,...\in \bar{B}$, for $\epsilon>0$, $\exists N$ such that $d(x_n,y_n)\leq d+\epsilon$ for $n\geq N$

Now each of the $x_i$ are in the $\bar{A}$, this means for each $x_i$, there exists $x_i'\in A$ such that $d(x_i,x_i')<\epsilon$ (In a metric space, the closure is the set of limit point of $A$, so there must be $d(x_i,x_i')<\epsilon$ with $x_i'\in A$). Similarly there exists $y_i'\in B$ such that $d(y_i,y_i')<\epsilon$. Then

$ d(\bar{A},\bar{B})\geq d(x_i,y_i)-\epsilon\geq d(x_i',y_i')-3\epsilon \geq d(A,B)-3\epsilon$

where I used $d(x_i,y_i)+d(x_i',x_i)+d(y_i,y_i')\geq d(x_i',y_i')$ but $\epsilon$ is arbitrary... so we are done.

Lost1
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  • I think is perfectly correct. – Ma Ming Feb 18 '14 at 14:03
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    @MaMing i think your profile photo is ironic because your surname is Ma. – Lost1 Feb 18 '14 at 14:04
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    Why do you use sequences, incidentally? Only one $x_1$, $y_1$ would be enough. – Clement C. Feb 18 '14 at 14:04
  • @ClementC. I am not saying this is a nice proof/shortest proof, but I think this is a reasonable proof. I am more annoyed about the attitude of the OP on the other post of needing further clarification on what I thought was a reasonable hint. Perhaps this should be a meta question. I am even more convinced that the OP on the other post didn't understand the alternative proof which I think is more complicated. – Lost1 Feb 18 '14 at 14:06

1 Answers1

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The proof is correct. Just to have a shorter proof of the same fact (the distance between sets is equal to the distance between their closures), here goes:

Let $\rho = d(A,B)$. The inequality $d(\overline{A},\overline{B})\le \rho$ holds because the infimum on the left is over a larger set. In the opposite direction, take any $a\in \overline{A}$ and $b\in \overline{B}$. The goal is to prove $d(a,b)\ge \rho$.

For every $\epsilon>0$ there exist $a'\in A$ and $b'\in B$ such that $d(a,a')<\epsilon$ and $d(b,b')<\epsilon$. By the triangle inequality, $\rho \le d(a',b')\le d(a,b)+ 2\epsilon$. Since $\epsilon$ can be arbitrarily small, $\rho\le d(a,b)$.